how to print intermediate result in Functor and Applicative in Haskell

Question

I am reading the book Programming in Haskell 2nd by Graham Hutton. https://www.cs.nott.ac.uk/~pszgmh/pih.html#slides

When it comes to chapter 13.4 Sequencing parser

Answer 1

It's a good start. Your last step doesn't look quite right. Stripping off the shared parts, you've written that

\inp -> case (\inp -> [(g,inp)]) inp of [] -> []; [(g, out)] -> parse (fmap g item) out
=
                                                                parse (fmap g item) out

This equation doesn't look quite right to me: the former is a function, and the latter isn't. Additionally, the latter refers to the free variable out, while the former doesn't (because out is bound by the pattern match it's enclosed in). A more careful continuation looks like this:

\inp -> case (\inp -> [(g,inp)]) inp of [] -> []; [(g, out)] -> parse (fmap g item) out
= { beta reduction, substituting inp for inp }
\inp -> case [(g, inp)]              of [] -> []; [(g, out)] -> parse (fmap g item) out
= { case reduction, substituting g for g and inp for out }
\inp ->                                                         parse (fmap g item) inp

If you like, you could then eta-reduce this to parse (fmap g item). Plugging this back into the shared parts we dropped above, we have:

three
=
P (parse (fmap g item)) <*> item <*> item

The result can be verified in ghci:

*Parsing> parse three "abc"
[(('a','c'),"")]

*Parsing> let g x y z = (x,z)

*Parsing> parse (P (parse (fmap g item)) <*> item <*> item) "abc"
[(('a','c'),"")]

As next steps, there are three places you could do your next definition-expansion to enable further progress:

1. You could expand the fmap in fmap g item.
2. You could expand the inner (<*>) in P (...) <*> item.
3. You could expand the outer (<*>) in (P (...) <*> item) <*> item.

Good luck!